Find all the second partial derivatives. v = e5xey

1 answer:

5 0

I'm taking the liberty of editing your function v = e5xey: It should be

v = e^5x^ey, with " ^ " indicating exponentiation.

Did you mean e^(5x) or (e^5)x? I'll assume it's e^(5x).

The partial of v = e^(5x)e^y with respect to x is e^(5x)(5)*e^y, or 25x*e^y.

The partial of v = e^(5x)e^y with respect to y is e^(5x)e^y.

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Write the equation of a line that passes a given point and has the given slope. Answer in standard form. 1. Point (-3,5), m = -7

Likurg_2 [28]

Answer:

y -5 = -7(x +3)
y = 7

Step-by-step explanation:

The point-slope form of the equation can be used for this. It is generally written as ...

y -k = m(x -h) . . . . . line with slope m through point (h, k)

1. For m=-7, (h, k) = (-3, 5), the equation is ...

y -5 = -7(x +3)

2. For m=0, (h, k) = (4, 7), the equation is ...

y -7 = 0(x -4)

y = 7

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Comment on the form of the answer

Since a point and slope are given, "the equation" is appropriately written in point-slope form. For the second problem the simplified version would be ...

y -7 = 0

Sometimes, "the equation" means the answer is expected in slope-intercept form. In that case, a little rearranging is necessary for the first answer. It becomes ...

y = -7x -16

There are perhaps a dozen different forms that a linear equation can be written in. Considering those, "the equation" is somewhat ambiguous. Your curriculum materials may offer a clue as to the form intended.

5 0

3 years ago

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I am Lyosha [343]

Answer:

(8, 15 and 17)

Step-by-step explanation:

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